If k is an integer greater than 6, all of the following must be divisi

Since we are working with VARIABLE, We can be sure if the product can be shown as 3 consecutive numbers, by adding /subtracting 3 from any of the terms..

lets see the choices

a) k(k+3)(k-1)
(k-1)*k*(k+1) are consecutives BUT instead of k+1, we have k+3 in k(k+3)(k-1) ..
and if k+1 is div by 3, k+3 will not and k and k-1 will also be not div by 3..
so this is NOT necessary to be div by 3

b)3k^3
3 is there in the term, so TRUE

c)(k+1)(k+5)(k+6)
if K+1 is div by 3, k+1+3=k+4 will also be div by 3
so we have consecutive terms by adding 3 to k+1
k+4, k+5 and k+6..
TRUE

d)(k+2)(k-2)(k+3)
add 3 to k-2. it becomes k-2+3=k+1..
again 3 consecutive terms k+1, k+2 and k+3
TRUE

e)k(k+1)(k+2)
3 consecutive terms already
TRUE
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Similar questions to practice:
if-n-is-an-integer-greater-than-6-which-of-the-following-56233.html
if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html
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I looked at this question and thought why would somebody make this question and post it here. then I saw that it is from GMATPREP. So now i assume this is one of those simple questions that make you think about your performance while doing the CAT. Algorithm's tricks.

I figured it out by taking K=7 and k=8
and option A is not divisible by 3 in case of k=8
so A is a answer.

Thanks !

I still don't understand this one why it would be choice A.

The question asks: If k is an integer greater than 6, then all the following must be divisible by 3 EXCEPT:

k(k+3)(k-1) (option A) is not necessarily divisible by 6. For example, if k = 8, then k(k+3)(k-1)=8*11*7, which is not divisible by 6. So, the answer is A.
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Got it now. Thank you!

When should one use plug in method to solve a question? Is this method of plugging in better?

k(k+3)(k-1) will not be divisible by 3 when k=3n-1, where n is any positive integer
A

Just test k=11.

Each answer choice but A will lead to a set of numbers divisible by 3.

FIrst, we eliminate choices B and E because both of those are divisible by 3 (3k^3 clearly has a factor of 3, and k(k+1)(k+2) is a product of 3 consecutive integers).

Now, let’s examine the other choices by letting k = 7.

A) k(k + 3)(k - 1) = 7(10)(6) is divisible by 3.

C) (k+1)(k+5)(k+6) = 8(12)(13) is divisible by 3.

D) (k+2)(k-2)(k+3) = 9(5)(10) is divisible by 3.

We see that all these choices are divisible by 3 when k = 7. We need to change k to a different number, so let k = 8.

A) k(k + 3)(k - 1) = 8(11)(7) is NOT divisible by 3.

We see that we’ve found the correct choice. However, let’s show that the other two indeed are divisible by 3.

C) (k+1)(k+5)(k+6) = 9(13)(14) is divisible by 3.

D) (k+2)(k-2)(k+3) = 10(6)(11) is divisible by 3.

Answer: A
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I am confused by this question. I used 10 as my "plug in" number since it said greater than 6, and 10 seemed like an easy number to keep track of. For answer A it works out to (10)(13)(9) which equals 1170, which is divisible by 3, unless my math is wrong.

What did I do wrong here?

This is a MUST be question.
Take k=11, then 11*14*10 is not divisible by 3
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How are following two consecutive

C. (k+1)(k+5)(k+6)
D. (k+2)(k-2)(k+3)

A. k(k + 3)(k - 1)

Take k=8,

Then, k(k + 3)(k - 1)=8x11X7 which is not divisible by 3, hence answer A